Question 13.22: In the circuit shown in Fig. 13.37, the source output resist...

Because the source resistance is larger than the load resistance, we associate the shunt capacitor with the source (to decrease the apparent source impedance) and the series inductor with the load (to increase the apparent load impedance). The modified source and load impedances are given by

\begin{aligned} &Z_L(j \omega)=R_L+j \omega L \\ &Z_o(j \omega)=\frac{R_o}{1+j \omega R_o C}, \omega=2 \pi f \end{aligned}

 For maximum power transfer at the match frequency f_0=60 Hz, we require Z_L\left(j \omega_0\right)=Z_o{ }^*\left(j \omega_0\right), which leads to

\frac{R_o\left(1-j \omega_0 R_o C\right)}{1+\left(\omega_0 R_o C\right)^2}=R_L-j \omega_0 L ; \omega_0=2 \pi f_0

It follows that

\begin{aligned} &\frac{R_o}{1+\left(\omega_0 R_o C\right)^2}=R_L \\ &\frac{R_o^2 C}{1+\left(\omega_0 R_o C\right)^2}=R_L R_o C=L . \end{aligned}

 Solving the first of these relations for the capacitance and then using the second to calculate the inductance yields

\begin{gathered} C=\frac{1}{\omega_0 R_o} \sqrt{\frac{R_0}{R_L}-1}=79.6 \mu F \\\\ L=R_o R_L C=79.6 mH \end{gathered}             (13.63)

The power delivered to the load (as a function of frequency) is given by

\begin{aligned} P_L(f) &=\frac{1}{2}|\tilde{I}(f)|^2 \operatorname{Re}\left[Z_L(j 2 \pi f)\right] \\ &=\frac{1}{2}\left|\frac{\tilde{V}(f)}{Z_o(j 2 \pi f)+Z_L(j 2 \pi f)}\right|^2 R_L . \end{aligned}

Figure  13.39  shows the power transferred to the load, normalized to the power transferred at the match frequency, as a function of frequency. The power transferred to the load is maximum at a frequency slightly higher than the match frequency f_0 . As an exercise, you can show that the frequency for maximum power transfer is f_1=65.8 Hz .